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## Millikan’s Oil Drop Experiment

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**Millikan’s Oil Drop Experiment**To Determine The Charge Of The Electron**Success Criteria**• Be able to state and explain Stokes Law for bodies moving through fluids • Be able to identify the forces acting on a stationary charged object in an electric field • Recall and explain how Millikan was experimentally able to determine the charge on the electron**Falling through a fluid**• Give a detailed description of the forces acting on an object moving in a fluid • How could the drag force on the object be calculated?**Measuring the viscosity of water**• Measurements: • Mass of ball bearing • Radius of ball bearing • Terminal velocity of ball bearing**Stoke’s Law**• When an object is dropped through a fluid, it experiences a force called viscous drag. • This force acts in the opposite direction to the velocity of the object, and is due to the viscosity of the fluid. • The force on a spherical object can be calculated using Stoke’s Law: F=6πηrv η=coeff of viscosity of fluid, r=radius of object v=velocity of object, π=Pi**Assumptions**• Small Reynolds number (small particles satisfy this) • Laminar flow • Smooth, spherical particles • Homogeneous fluid • Particles do not interfere with each other**Millikan’s Apparatus**• Atomiser created a fine mist of oil droplets, charged by friction (Later experiments used x-rays). • Some droplets fell through hole and could be viewed through the microscope with a scale to measure distances and hence velocities. • Millikan applied an electric field between the plates which would exert a force on the droplets. He could adjust the p.d. and hence field strength to get the oil droplets to hover.**Forces without a field**Before the electric field is turned on, the forces on the droplets are: a) the weight of the drop; b) the viscous force from the air. The drop will reach terminal velocity when the two forces are equal. mg=6πηrv The mass of the drop is the volume multiplied by the density of the oil, ρ , so the equation can be written: 4/3πr3ρg = 6πηrv => r2 = 9ηv/2ρg Millikan in earlier experiments measured η and ρ, and now can calculate r. Viscous Force Weight**With An Electric Field**Millikan adjusted p.d. till the drop was stationary. Since for viscous force to act the drop needed to be moving, this for is now not there. Two forces now are: • Weight of droplet; • Force due to uniform electric field.**For a charged oil droplet**• The electric force is given by: • F = QV/d Where • Q=charge on oil drop • V=p.d. between plates • d=distance between plates**If drop is stationary then electric force must be equal to**the weight. QV/d = 4/3πr3ρg Earlier, r was calculated so the only unknown is Q. Millikan could calculate the charge on the droplet. He repeated his experiment many times and found that the smallest value calculated was -1.6X10-19 C and all other values were a multiple of it. He concluded that charge could never exist in smaller quantities than this, and this was the charge carried by a single electron.**Practical Task**• Find the coefficient of viscosity of glycerine